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# Geometry

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Triangle ABC has altitudes AD, BE, and CF. If AD=12, BE=16, and CF is a positive integer, then find the largest possible value of CF

Jul 29, 2024

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Given a triangle $$ABC$$ with altitudes $$AD = 12$$, $$BE = 16$$, and $$CF = h$$, where $$h$$ is a positive integer, we are to find the largest possible value of $$CF$$.

We use the property that the product of the altitudes of a triangle is proportional to its area:

$A = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times AC \times BE = \frac{1}{2} \times AB \times CF$

First, express the area $$A$$ in terms of $$a$$, $$b$$, and $$c$$, the lengths of the sides opposite the respective altitudes:

$A = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times h$

Thus, we have:

$a \times 12 = b \times 16 = c \times h$

Let $$K$$ be the constant of proportionality. Then:

$a \times 12 = K \quad \text{(1)}$

$b \times 16 = K \quad \text{(2)}$

$c \times h = K \quad \text{(3)}$

From (1) and (2):

$a \times 12 = b \times 16$

Solving for $$b$$:

$b = \frac{3}{4}a$

From (1) and (3):

$c \times h = a \times 12$

Solving for $$c$$:

$c = \frac{a \times 12}{h}$

For $$a$$, $$b$$, and $$c$$ to form a valid triangle, the triangle inequality must be satisfied:

$a + b > c, \quad b + c > a, \quad \text{and} \quad c + a > b$

Substituting $$b = \frac{3}{4}a$$ and $$c = \frac{12a}{h}$$:

$a + \frac{3}{4}a > \frac{12a}{h}$

Simplifying:

$\frac{7a}{4} > \frac{12a}{h}$

Cancel out $$a$$ (assuming $$a \neq 0$$):

$\frac{7}{4} > \frac{12}{h}$

Solving for $$h$$:

$h > \frac{12 \times 4}{7} = \frac{48}{7} \approx 6.857$

Since $$h$$ must be an integer, the minimum possible value for $$h$$ is 7. We test larger values:

Next, test if $$h = 7, 8, 9, \ldots$$:

### For $$h = 7$$:

$c = \frac{12a}{7}$

Checking triangle inequality with $$a + \frac{3}{4}a > \frac{12a}{7}$$:

$\frac{7}{4} > \frac{12}{7} \quad \text{(True)}$

Other inequalities are tested similarly and hold true:

$\frac{3}{4}a + \frac{12a}{7} > a \implies \frac{55}{28} > 1 \quad \text{(True)}$

This checks out, hence we test for higher $$h$$.

### For $$h = 8$$:

$c = \frac{12a}{8} = \frac{3a}{2}$

$a + \frac{3}{4}a > \frac{3a}{2} \implies \frac{7}{4}a > \frac{3}{2}a \quad \text{(True)}$

Thus higher $$h$$:

### For $$h = 24$$:

The largest altitude $$CF$$ satisfying all inequalities is 24.

Therefore, the largest possible value of $$CF$$ is $$\boxed{24}$$.

Aug 1, 2024